(if anyone in AMP UCSC is reading this, it will sound very very familiar ) So let's just transfer this chessboard to an infinite lattice grid, and the condition states that cardboard piece must have sides parallel to each axis and have each vertex be at a lattice point. Tile the plane with $n\times n$ squares, so that each of their vertices are at $(kn, ln)$ for all integers $k, l$. Now, let's "mod" each of these squares to the square with vertices at $(0,0), (n,0), (0,n),$ and $(n,n)$, so that each unit square with lower-left vertex $(a,b)$ will go to the unit square with lower-left vertex $(a\pmod{n}, b\pmod{n})$. Note that each cardboard square, when placed on the grid and "modded" to this central nxn square, will cover every point in this nxn square exactly once. Therefore, each unit square with lower-left vertex $(x,y)$ with $0\le x<n$, $0\le y<n$ will be fully covered exactly 2009 times. Now, we claim that before the modding, among all of the unit squares with lower-left vertices $(x+kn, y+ln)$ for all integers $k, l$, at least one of them is covered by an odd number of cardboard pieces. If each of them were covered by an even number of pieces, then after the modding, the unit square with lower-left vertex $(x,y)$ would be covered an even number of times, a contradiction as 2009 is an odd number. Therefore, for all $x$ and $y$ such that $0\le x<n$, $0\le y<n$, there exists a unit square with lower-left vertex $(x+kn, y+ln)$ that is covered by an odd number of pieces. As there are $n^2$ possible combinations of $x$ and $y$, and each of these gives at least one unique unit square that is covered an odd number of times, there are at least $n^2$ unit squares that are covered by an odd number of pieces, which was what was wanted.